Logarithmic and exponential functions — practice questions

This equation is a quadratic in disguise. Let $y = e^x$. Then the equation becomes:

$y^2 - 7y + 10 = 0$ [M1 for substitution or recognising quadratic form]

Factorising the quadratic: $(y-5)(y-2) = 0$

So, $y=5$ or $y=2$. [A1 for correct values of y]

Now substitute back $e^x$ for $y$: Case 1: $e^x = 5$ Taking natural logarithms of both sides: $\ln(e^x) = \ln 5$ $x = \ln 5$ [A1]

Case 2: $e^x = 2$ Taking natural logarithms of both sides: $\ln(e^x) = \ln 2$ $x = \ln 2$ [A1]

The solutions are $x = \ln 2$ and $x = \ln 5$.

(i) First, we need to find the derivative, $\frac{dy}{dx}$. Using the chain rule for $\ln(f(x))$ with $f(x) = x^2+4$, we have $f'(x) = 2x$. So, $\frac{dy}{dx} = \frac{f'(x)}{f(x)} = \frac{2x}{x^2+4}$. [M1 for correct application of chain rule]

At $x=2$, the gradient is: $\frac{dy}{dx} = \frac{2(2)}{2^2+4} = \frac{4}{4+4} = \frac{4}{8} = \frac{1}{2}$. [A1]

(ii) Stationary points occur when $\frac{dy}{dx} = 0$. $\frac{2x}{x^2+4} = 0$ [M1 for setting derivative to 0] For a fraction to be zero, the numerator must be zero. $2x = 0 \implies x = 0$.

To find the y-coordinate, substitute $x=0$ back into the original equation: $y = \ln(0^2 + 4) = \ln 4$.

The exact coordinates of the stationary point are $(0, \ln 4)$. [A1 for both coordinates]

(i) The initial mass occurs at $t=0$. $M = 200e^{-0.05(0)} = 200e^0 = 200 \times 1 = 200$ grams. [B1]

(ii) After 10 years, $t=10$. $M = 200e^{-0.05(10)} = 200e^{-0.5}$. $M \approx 200 \times 0.60653... \approx 121.306...$ The mass is 121 grams (to 3 s.f.). [M1 for substitution, A1 for correct answer]

(iii) The initial mass is 200g, so half the mass is 100g. We need to find $t$ when $M=100$. $100 = 200e^{-0.05t}$ [M1 for setting up the equation] Divide by 200: $0.5 = e^{-0.05t}$. Take natural logarithms of both sides: $\ln(0.5) = \ln(e^{-0.05t})$ $\ln(0.5) = -0.05t$ [M1 for taking logs correctly] Now, solve for $t$: $t = \frac{\ln(0.5)}{-0.05} = \frac{-0.6931...}{-0.05} \approx 13.862...$ The time taken is approximately 13.9 years. [A1]